Integrand size = 86, antiderivative size = 29 \[ \int \frac {\left (-405+2 x+x^2\right ) \log \left (\frac {1}{2} \left (405-2 x-x^2\right )\right )+\log (x) \left (2 x+2 x^2+\left (405-2 x-x^2\right ) \log \left (\frac {1}{2} \left (405-2 x-x^2\right )\right )\right )}{\left (-810 x^2+4 x^3+2 x^4\right ) \log (2)} \, dx=\frac {\log (x) \log \left (-x+\frac {1}{2} \left (405-x^2\right )\right )}{2 x \log (2)} \]
Time = 0.40 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-405+2 x+x^2\right ) \log \left (\frac {1}{2} \left (405-2 x-x^2\right )\right )+\log (x) \left (2 x+2 x^2+\left (405-2 x-x^2\right ) \log \left (\frac {1}{2} \left (405-2 x-x^2\right )\right )\right )}{\left (-810 x^2+4 x^3+2 x^4\right ) \log (2)} \, dx=\frac {\log (x) \log \left (\frac {1}{2} \left (405-2 x-x^2\right )\right )}{2 x \log (2)} \]
Integrate[((-405 + 2*x + x^2)*Log[(405 - 2*x - x^2)/2] + Log[x]*(2*x + 2*x ^2 + (405 - 2*x - x^2)*Log[(405 - 2*x - x^2)/2]))/((-810*x^2 + 4*x^3 + 2*x ^4)*Log[2]),x]
Time = 1.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {27, 27, 2026, 7279, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (x^2+2 x-405\right ) \log \left (\frac {1}{2} \left (-x^2-2 x+405\right )\right )+\log (x) \left (2 x^2+\left (-x^2-2 x+405\right ) \log \left (\frac {1}{2} \left (-x^2-2 x+405\right )\right )+2 x\right )}{\left (2 x^4+4 x^3-810 x^2\right ) \log (2)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\left (-x^2-2 x+405\right ) \log \left (\frac {1}{2} \left (-x^2-2 x+405\right )\right )-\log (x) \left (2 x^2+2 x+\left (-x^2-2 x+405\right ) \log \left (\frac {1}{2} \left (-x^2-2 x+405\right )\right )\right )}{2 \left (-x^4-2 x^3+405 x^2\right )}dx}{\log (2)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\left (-x^2-2 x+405\right ) \log \left (\frac {1}{2} \left (-x^2-2 x+405\right )\right )-\log (x) \left (2 x^2+2 x+\left (-x^2-2 x+405\right ) \log \left (\frac {1}{2} \left (-x^2-2 x+405\right )\right )\right )}{-x^4-2 x^3+405 x^2}dx}{2 \log (2)}\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \frac {\int \frac {\left (-x^2-2 x+405\right ) \log \left (\frac {1}{2} \left (-x^2-2 x+405\right )\right )-\log (x) \left (2 x^2+2 x+\left (-x^2-2 x+405\right ) \log \left (\frac {1}{2} \left (-x^2-2 x+405\right )\right )\right )}{x^2 \left (-x^2-2 x+405\right )}dx}{2 \log (2)}\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \frac {\int \left (\frac {2 (x+1) \log (x)}{x \left (x^2+2 x-405\right )}-\frac {(\log (x)-1) \log \left (\frac {1}{2} \left (-x^2-2 x+405\right )\right )}{x^2}\right )dx}{2 \log (2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\log (x) \log \left (\frac {1}{2} \left (-x^2-2 x+405\right )\right )}{2 x \log (2)}\) |
Int[((-405 + 2*x + x^2)*Log[(405 - 2*x - x^2)/2] + Log[x]*(2*x + 2*x^2 + ( 405 - 2*x - x^2)*Log[(405 - 2*x - x^2)/2]))/((-810*x^2 + 4*x^3 + 2*x^4)*Lo g[2]),x]
3.20.67.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.44 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {\ln \left (x \right ) \ln \left (-\frac {1}{2} x^{2}-x +\frac {405}{2}\right )}{2 \ln \left (2\right ) x}\) | \(23\) |
parallelrisch | \(\frac {\ln \left (x \right ) \ln \left (-\frac {1}{2} x^{2}-x +\frac {405}{2}\right )}{2 \ln \left (2\right ) x}\) | \(23\) |
default | \(\frac {-\frac {\ln \left (-x^{2}-2 x +405\right )}{x}+\frac {\ln \left (x^{2}+2 x -405\right )}{405}-\frac {2 \sqrt {406}\, \operatorname {arctanh}\left (\frac {\left (2+2 x \right ) \sqrt {406}}{812}\right )}{405}+\frac {\left (\ln \left (x \right )+1\right ) \ln \left (-x^{2}-2 x +405\right )}{x}-\frac {\ln \left (x +1-\sqrt {406}\right )}{405}-\frac {\ln \left (x +1-\sqrt {406}\right ) \sqrt {406}}{405}-\frac {\ln \left (1+\sqrt {406}+x \right )}{405}+\frac {\ln \left (1+\sqrt {406}+x \right ) \sqrt {406}}{405}-\frac {\ln \left (2\right ) \ln \left (x \right )}{x}}{2 \ln \left (2\right )}\) | \(125\) |
parts | \(-\frac {\ln \left (x \right )}{2 x}-\frac {1}{2 x}+\frac {\frac {\left (\ln \left (x \right )+1\right ) \ln \left (-x^{2}-2 x +405\right )}{x}+\frac {2 \ln \left (x \right )}{405}-\frac {\ln \left (x +1-\sqrt {406}\right )}{405}-\frac {\ln \left (x +1-\sqrt {406}\right ) \sqrt {406}}{405}-\frac {\ln \left (1+\sqrt {406}+x \right )}{405}+\frac {\ln \left (1+\sqrt {406}+x \right ) \sqrt {406}}{405}}{2 \ln \left (2\right )}+\frac {-\frac {\ln \left (-x^{2}-2 x +405\right )}{x}-\frac {2 \ln \left (x \right )}{405}+\frac {\ln \left (x^{2}+2 x -405\right )}{405}-\frac {2 \sqrt {406}\, \operatorname {arctanh}\left (\frac {\left (2+2 x \right ) \sqrt {406}}{812}\right )}{405}+\frac {\ln \left (2\right )}{x}}{2 \ln \left (2\right )}\) | \(150\) |
int((((-x^2-2*x+405)*ln(-1/2*x^2-x+405/2)+2*x^2+2*x)*ln(x)+(x^2+2*x-405)*l n(-1/2*x^2-x+405/2))/(2*x^4+4*x^3-810*x^2)/ln(2),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {\left (-405+2 x+x^2\right ) \log \left (\frac {1}{2} \left (405-2 x-x^2\right )\right )+\log (x) \left (2 x+2 x^2+\left (405-2 x-x^2\right ) \log \left (\frac {1}{2} \left (405-2 x-x^2\right )\right )\right )}{\left (-810 x^2+4 x^3+2 x^4\right ) \log (2)} \, dx=\frac {\log \left (-\frac {1}{2} \, x^{2} - x + \frac {405}{2}\right ) \log \left (x\right )}{2 \, x \log \left (2\right )} \]
integrate((((-x^2-2*x+405)*log(-1/2*x^2-x+405/2)+2*x^2+2*x)*log(x)+(x^2+2* x-405)*log(-1/2*x^2-x+405/2))/(2*x^4+4*x^3-810*x^2)/log(2),x, algorithm=\
Time = 0.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {\left (-405+2 x+x^2\right ) \log \left (\frac {1}{2} \left (405-2 x-x^2\right )\right )+\log (x) \left (2 x+2 x^2+\left (405-2 x-x^2\right ) \log \left (\frac {1}{2} \left (405-2 x-x^2\right )\right )\right )}{\left (-810 x^2+4 x^3+2 x^4\right ) \log (2)} \, dx=\frac {\log {\left (x \right )} \log {\left (- \frac {x^{2}}{2} - x + \frac {405}{2} \right )}}{2 x \log {\left (2 \right )}} \]
integrate((((-x**2-2*x+405)*ln(-1/2*x**2-x+405/2)+2*x**2+2*x)*ln(x)+(x**2+ 2*x-405)*ln(-1/2*x**2-x+405/2))/(2*x**4+4*x**3-810*x**2)/ln(2),x)
Time = 0.33 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {\left (-405+2 x+x^2\right ) \log \left (\frac {1}{2} \left (405-2 x-x^2\right )\right )+\log (x) \left (2 x+2 x^2+\left (405-2 x-x^2\right ) \log \left (\frac {1}{2} \left (405-2 x-x^2\right )\right )\right )}{\left (-810 x^2+4 x^3+2 x^4\right ) \log (2)} \, dx=-\frac {\log \left (2\right ) \log \left (x\right ) - \log \left (-x^{2} - 2 \, x + 405\right ) \log \left (x\right )}{2 \, x \log \left (2\right )} \]
integrate((((-x^2-2*x+405)*log(-1/2*x^2-x+405/2)+2*x^2+2*x)*log(x)+(x^2+2* x-405)*log(-1/2*x^2-x+405/2))/(2*x^4+4*x^3-810*x^2)/log(2),x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {\left (-405+2 x+x^2\right ) \log \left (\frac {1}{2} \left (405-2 x-x^2\right )\right )+\log (x) \left (2 x+2 x^2+\left (405-2 x-x^2\right ) \log \left (\frac {1}{2} \left (405-2 x-x^2\right )\right )\right )}{\left (-810 x^2+4 x^3+2 x^4\right ) \log (2)} \, dx=-\frac {\frac {\log \left (2\right ) \log \left (x\right )}{x} - \frac {\log \left (-x^{2} - 2 \, x + 405\right ) \log \left (x\right )}{x}}{2 \, \log \left (2\right )} \]
integrate((((-x^2-2*x+405)*log(-1/2*x^2-x+405/2)+2*x^2+2*x)*log(x)+(x^2+2* x-405)*log(-1/2*x^2-x+405/2))/(2*x^4+4*x^3-810*x^2)/log(2),x, algorithm=\
Time = 10.66 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {\left (-405+2 x+x^2\right ) \log \left (\frac {1}{2} \left (405-2 x-x^2\right )\right )+\log (x) \left (2 x+2 x^2+\left (405-2 x-x^2\right ) \log \left (\frac {1}{2} \left (405-2 x-x^2\right )\right )\right )}{\left (-810 x^2+4 x^3+2 x^4\right ) \log (2)} \, dx=\frac {\ln \left (-\frac {x^2}{2}-x+\frac {405}{2}\right )\,\ln \left (x\right )}{2\,x\,\ln \left (2\right )} \]